Question: $f\,^{\prime}(x)=8x^3-12x$ and $f(1)=1$. $f(-2) = $
Explanation: Finding $f(x)$ We have $f'(x)=8x^3-12x$ and we want to find $f(x)$ : $\begin{aligned}f(x) &= \int f'(x)\,dx \\\\ &= \int (8x^3-12x)\,dx \\\\ & = {2x^4-6x^2} {+ C} \end{aligned}$ Finding $ C$ Goal: We need to find $ C$ such that $f(1)=1$. Here's what we get when we plug in $1$ : $\begin{aligned}f(1)&={2(1)^4-6(1)^2} {+ C}\\\\ &={-4} {+ C} \end{aligned}$ We are given that this must equal $1$ : $1 = {-4} {+ C}$ Solving the equation gives us ${C=5}$. Finding $f(-2)$ Now, we have that $f(x)={2x^4-6x^2} {+5}$. Let's find $f(-2)$ by plugging in $-2$ : $\begin{aligned}f(-2)&=2(-2)^4-6(-2)^2+5\\\\ &=13 \end{aligned}$ The answer $f(-2) = 13$